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Electromagnetic Wave

155599 Among the following, the incorrect Maxwells Electromagnetic equation is

1

\int \vec{B} \cdot d \vec{l} = μ_{o} i_{c} + μ_{o} ε_{o} \frac{d ϕ_{E}}{dt}

2

f \vec{B} \cdot d \vec{A} = \frac{Q}{ε_{o}}

3

\int \vec{E} \cdot d \vec{ℓ} = \frac{- d ϕ_{B}}{dt}

4

f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{o}}

Explanation:

B We know that, the four Maxwell's equation are -
(i) $\int \vec{B} \cdot d \vec{A} = 0$
(ii) $f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{0}}$
(iii) $f \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{dt}$
(iv) $f \vec{B} \cdot d \vec{l} = μ_{0} (i_{c} + ε_{0} \frac{d ϕ_{E}}{dt})$
Option (b) is incorrect Maxwell equation.

AP EAMCET-06.07.2022

Electromagnetic Wave

155600 A plane electromagnetic wave of frequency 50 $MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively, then the correct option in the following is

1

K_{E} = K_{B}

2

K_{E} = K_{B} = 0

3

K_{E} > K_{B}

4

K_{E} < K_{B}

Explanation:

A :We know that,
$K_{E} = \frac{1}{2} \in_{0} E^{2}$
$E = cB$
$c^{2} = \frac{1}{μ_{0} \in_{0}}$
$K_{E} = \frac{1}{2} \in_{0} \frac{1}{μ_{0} \in_{0}} \times B^{2} = \frac{B^{2}}{2 μ_{0}} = K_{B}$
$K_{E} = K_{B}$

AP EAMCET-05.07.2022

Electromagnetic Wave

155601 The electric field (E) and magnetic field (B) of an electromagnetic wave passing through vacuum are given by
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$
Then the correct statement among the following is
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$

1

E_{0} k = B_{0} ω

2

E_{o} ω = B_{0} k

3

E_{0} B_{o} = ω k

4

E_{o} B_{o} = \frac{ω}{k}

Explanation:

A The relation between electric field and magnetic field is,
$\frac{E_{o}}{B_{o}} = c \Rightarrow E_{o} = {cB}_{0}$
And, $c = v λ = \frac{ω}{2 π} λ = \frac{ω}{k} {∵ k = \frac{2 π}{λ}, \frac{1}{k} = \frac{λ}{2 π}}$
$E_{o} = B_{o} c$
$E_{o} = B_{0} \times \frac{ω}{k}$
$E_{o} k = B_{o} ω$

JEE Main-24.01.2023

Electromagnetic Wave

155602 The ratio of amplitudes of electric and magnetic fields i.e. $\frac{E_{0}}{B_{0}}$ is equal to

1

\sqrt{μ_{0} ε_{0}}

2

\frac{1}{\sqrt{μ_{0} ε_{0}}}

3

μ_{0} ε_{0}

4

\frac{1}{μ_{0} ε_{0}}

Explanation:

B Given that,
$\frac{E_{0}}{B_{0}} = c$
$\frac{| E |}{| H |} = η = \sqrt{\frac{μ}{ε}}$
Since, $B = μ H$
$H = \frac{B}{μ}$
$\frac{| E_{0} |}{| B_{0} |} μ = \sqrt{\frac{μ}{ε}}$
$\frac{| E_{0} |}{| B_{0} |} = \sqrt{\frac{1}{μ ε}}$
$c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

Assam CEE-31.07.2022

Electromagnetic Wave

155599 Among the following, the incorrect Maxwells Electromagnetic equation is

1

\int \vec{B} \cdot d \vec{l} = μ_{o} i_{c} + μ_{o} ε_{o} \frac{d ϕ_{E}}{dt}

2

f \vec{B} \cdot d \vec{A} = \frac{Q}{ε_{o}}

3

\int \vec{E} \cdot d \vec{ℓ} = \frac{- d ϕ_{B}}{dt}

4

f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{o}}

Explanation:

B We know that, the four Maxwell's equation are -
(i) $\int \vec{B} \cdot d \vec{A} = 0$
(ii) $f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{0}}$
(iii) $f \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{dt}$
(iv) $f \vec{B} \cdot d \vec{l} = μ_{0} (i_{c} + ε_{0} \frac{d ϕ_{E}}{dt})$
Option (b) is incorrect Maxwell equation.

AP EAMCET-06.07.2022

Electromagnetic Wave

155600 A plane electromagnetic wave of frequency 50 $MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively, then the correct option in the following is

1

K_{E} = K_{B}

2

K_{E} = K_{B} = 0

3

K_{E} > K_{B}

4

K_{E} < K_{B}

Explanation:

A :We know that,
$K_{E} = \frac{1}{2} \in_{0} E^{2}$
$E = cB$
$c^{2} = \frac{1}{μ_{0} \in_{0}}$
$K_{E} = \frac{1}{2} \in_{0} \frac{1}{μ_{0} \in_{0}} \times B^{2} = \frac{B^{2}}{2 μ_{0}} = K_{B}$
$K_{E} = K_{B}$

AP EAMCET-05.07.2022

Electromagnetic Wave

155601 The electric field (E) and magnetic field (B) of an electromagnetic wave passing through vacuum are given by
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$
Then the correct statement among the following is
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$

1

E_{0} k = B_{0} ω

2

E_{o} ω = B_{0} k

3

E_{0} B_{o} = ω k

4

E_{o} B_{o} = \frac{ω}{k}

Explanation:

A The relation between electric field and magnetic field is,
$\frac{E_{o}}{B_{o}} = c \Rightarrow E_{o} = {cB}_{0}$
And, $c = v λ = \frac{ω}{2 π} λ = \frac{ω}{k} {∵ k = \frac{2 π}{λ}, \frac{1}{k} = \frac{λ}{2 π}}$
$E_{o} = B_{o} c$
$E_{o} = B_{0} \times \frac{ω}{k}$
$E_{o} k = B_{o} ω$

JEE Main-24.01.2023

Electromagnetic Wave

155602 The ratio of amplitudes of electric and magnetic fields i.e. $\frac{E_{0}}{B_{0}}$ is equal to

1

\sqrt{μ_{0} ε_{0}}

2

\frac{1}{\sqrt{μ_{0} ε_{0}}}

3

μ_{0} ε_{0}

4

\frac{1}{μ_{0} ε_{0}}

Explanation:

B Given that,
$\frac{E_{0}}{B_{0}} = c$
$\frac{| E |}{| H |} = η = \sqrt{\frac{μ}{ε}}$
Since, $B = μ H$
$H = \frac{B}{μ}$
$\frac{| E_{0} |}{| B_{0} |} μ = \sqrt{\frac{μ}{ε}}$
$\frac{| E_{0} |}{| B_{0} |} = \sqrt{\frac{1}{μ ε}}$
$c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

Assam CEE-31.07.2022

Electromagnetic Wave

155599 Among the following, the incorrect Maxwells Electromagnetic equation is

1

\int \vec{B} \cdot d \vec{l} = μ_{o} i_{c} + μ_{o} ε_{o} \frac{d ϕ_{E}}{dt}

2

f \vec{B} \cdot d \vec{A} = \frac{Q}{ε_{o}}

3

\int \vec{E} \cdot d \vec{ℓ} = \frac{- d ϕ_{B}}{dt}

4

f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{o}}

Explanation:

B We know that, the four Maxwell's equation are -
(i) $\int \vec{B} \cdot d \vec{A} = 0$
(ii) $f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{0}}$
(iii) $f \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{dt}$
(iv) $f \vec{B} \cdot d \vec{l} = μ_{0} (i_{c} + ε_{0} \frac{d ϕ_{E}}{dt})$
Option (b) is incorrect Maxwell equation.

AP EAMCET-06.07.2022

Electromagnetic Wave

155600 A plane electromagnetic wave of frequency 50 $MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively, then the correct option in the following is

1

K_{E} = K_{B}

2

K_{E} = K_{B} = 0

3

K_{E} > K_{B}

4

K_{E} < K_{B}

Explanation:

A :We know that,
$K_{E} = \frac{1}{2} \in_{0} E^{2}$
$E = cB$
$c^{2} = \frac{1}{μ_{0} \in_{0}}$
$K_{E} = \frac{1}{2} \in_{0} \frac{1}{μ_{0} \in_{0}} \times B^{2} = \frac{B^{2}}{2 μ_{0}} = K_{B}$
$K_{E} = K_{B}$

AP EAMCET-05.07.2022

Electromagnetic Wave

155601 The electric field (E) and magnetic field (B) of an electromagnetic wave passing through vacuum are given by
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$
Then the correct statement among the following is
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$

1

E_{0} k = B_{0} ω

2

E_{o} ω = B_{0} k

3

E_{0} B_{o} = ω k

4

E_{o} B_{o} = \frac{ω}{k}

Explanation:

A The relation between electric field and magnetic field is,
$\frac{E_{o}}{B_{o}} = c \Rightarrow E_{o} = {cB}_{0}$
And, $c = v λ = \frac{ω}{2 π} λ = \frac{ω}{k} {∵ k = \frac{2 π}{λ}, \frac{1}{k} = \frac{λ}{2 π}}$
$E_{o} = B_{o} c$
$E_{o} = B_{0} \times \frac{ω}{k}$
$E_{o} k = B_{o} ω$

JEE Main-24.01.2023

Electromagnetic Wave

155602 The ratio of amplitudes of electric and magnetic fields i.e. $\frac{E_{0}}{B_{0}}$ is equal to

1

\sqrt{μ_{0} ε_{0}}

2

\frac{1}{\sqrt{μ_{0} ε_{0}}}

3

μ_{0} ε_{0}

4

\frac{1}{μ_{0} ε_{0}}

Explanation:

B Given that,
$\frac{E_{0}}{B_{0}} = c$
$\frac{| E |}{| H |} = η = \sqrt{\frac{μ}{ε}}$
Since, $B = μ H$
$H = \frac{B}{μ}$
$\frac{| E_{0} |}{| B_{0} |} μ = \sqrt{\frac{μ}{ε}}$
$\frac{| E_{0} |}{| B_{0} |} = \sqrt{\frac{1}{μ ε}}$
$c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

Assam CEE-31.07.2022

Electromagnetic Wave

155599 Among the following, the incorrect Maxwells Electromagnetic equation is

1

\int \vec{B} \cdot d \vec{l} = μ_{o} i_{c} + μ_{o} ε_{o} \frac{d ϕ_{E}}{dt}

2

f \vec{B} \cdot d \vec{A} = \frac{Q}{ε_{o}}

3

\int \vec{E} \cdot d \vec{ℓ} = \frac{- d ϕ_{B}}{dt}

4

f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{o}}

Explanation:

B We know that, the four Maxwell's equation are -
(i) $\int \vec{B} \cdot d \vec{A} = 0$
(ii) $f \vec{E} \cdot d \vec{A} = \frac{Q}{ε_{0}}$
(iii) $f \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{dt}$
(iv) $f \vec{B} \cdot d \vec{l} = μ_{0} (i_{c} + ε_{0} \frac{d ϕ_{E}}{dt})$
Option (b) is incorrect Maxwell equation.

AP EAMCET-06.07.2022

Electromagnetic Wave

155600 A plane electromagnetic wave of frequency 50 $MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively, then the correct option in the following is

1

K_{E} = K_{B}

2

K_{E} = K_{B} = 0

3

K_{E} > K_{B}

4

K_{E} < K_{B}

Explanation:

A :We know that,
$K_{E} = \frac{1}{2} \in_{0} E^{2}$
$E = cB$
$c^{2} = \frac{1}{μ_{0} \in_{0}}$
$K_{E} = \frac{1}{2} \in_{0} \frac{1}{μ_{0} \in_{0}} \times B^{2} = \frac{B^{2}}{2 μ_{0}} = K_{B}$
$K_{E} = K_{B}$

AP EAMCET-05.07.2022

Electromagnetic Wave

155601 The electric field (E) and magnetic field (B) of an electromagnetic wave passing through vacuum are given by
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$
Then the correct statement among the following is
$E = E_{0} \sin (k x - ω t)$
$B = B_{0} \sin (k x - ω t)$

1

E_{0} k = B_{0} ω

2

E_{o} ω = B_{0} k

3

E_{0} B_{o} = ω k

4

E_{o} B_{o} = \frac{ω}{k}

Explanation:

A The relation between electric field and magnetic field is,
$\frac{E_{o}}{B_{o}} = c \Rightarrow E_{o} = {cB}_{0}$
And, $c = v λ = \frac{ω}{2 π} λ = \frac{ω}{k} {∵ k = \frac{2 π}{λ}, \frac{1}{k} = \frac{λ}{2 π}}$
$E_{o} = B_{o} c$
$E_{o} = B_{0} \times \frac{ω}{k}$
$E_{o} k = B_{o} ω$

JEE Main-24.01.2023

Electromagnetic Wave

155602 The ratio of amplitudes of electric and magnetic fields i.e. $\frac{E_{0}}{B_{0}}$ is equal to

1

\sqrt{μ_{0} ε_{0}}

2

\frac{1}{\sqrt{μ_{0} ε_{0}}}

3

μ_{0} ε_{0}

4

\frac{1}{μ_{0} ε_{0}}

Explanation:

B Given that,
$\frac{E_{0}}{B_{0}} = c$
$\frac{| E |}{| H |} = η = \sqrt{\frac{μ}{ε}}$
Since, $B = μ H$
$H = \frac{B}{μ}$
$\frac{| E_{0} |}{| B_{0} |} μ = \sqrt{\frac{μ}{ε}}$
$\frac{| E_{0} |}{| B_{0} |} = \sqrt{\frac{1}{μ ε}}$
$c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

Assam CEE-31.07.2022